Projectile Motion. Honors Physics
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1 Projectile Motion Honors Physics
2 What is projectile? Projectile -Any object which projected by some means and continues to moe due to its own inertia (mass).
3 Projectiles moe in TWO dimensions Since a projectile moes in - dimensions, it therefore has components just like a resultant ector. Horizontal and Vertical
4 Horizontal Velocity Component NEVER changes, coers equal displacements in equal time periods. This means the initial horizontal elocity equals the final horizontal elocity In other words, the horizontal elocity is CONSTANT. BUT WHY? Graity DOES NOT work horizontally to increase or decrease the elocity.
5 Vertical Velocity Component Changes (due to graity), does NOT coer equal displacements in equal time periods. Both the MAGNITUDE and DIRECTION change. As the projectile moes up the MAGNITUDE DECREASES and its direction is UPWARD. As it moes down the MAGNITUDE INCREASES and the direction is DOWNWARD.
6 Combining the Components Together, these components produce what is called a trajectory or path. This path is parabolic in nature. Component Horizontal Vertical Magnitude Constant Changes Direction Constant Changes
7 Horizontally Launched Projectiles Projectiles which hae NO upward trajectory and NO initial VERTICAL elocity. = = x constant = 0 m / s oy
8 Horizontally Launched Projectiles To analyze a projectile in dimensions we need equations. One for the x direction and one for the y direction. And for this we use kinematic #. x = t + 1 at x = t y = 1 gt Remember, the elocity is CONSTANT horizontally, so that means the acceleration is ZERO! Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO.
9 Horizontally Launched Projectiles Example: A plane traeling with a horizontal elocity of 100 m/s is 500 m aboe the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land? y = 1 gt 500 = 1 ( 9.8) t = t t = 10.1 seconds What do I know? =100 m/s y = 500 m oy = 0 m/s g = -9.8 m/s/s What I want to know? t =? x =? x = t = (100)(10.1) = 1010 m
10 Vertically Launched Projectiles NO Vertical Velocity at the top of the trajectory. Vertical Velocity decreases on the way upward Horizontal Velocity is constant Vertical Velocity increases on the way down, Component Horizontal Vertical Magnitude Constant Decreases up, top, Increases down Direction Constant Changes
11 Vertically Launched Projectiles Since the projectile was launched at a angle, the elocity MUST be broken into components!!! = o cos θ o oy oy = o sin θ θ
12 Vertically Launched Projectiles There are seeral things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground leel, the y displacement is ZERO: y = 0
13 Vertically Launched Projectiles You will still use kinematic #, but YOU MUST use COMPONENTS in the equation. o oy x = t y = 1 oyt + gt θ oy = = o o cosθ sinθ
14 Example A place kicker kicks a football with a elocity of 0.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? (b) How far away does it land? (c) How high does it trael? o =0.0 m/s θ = 53 oy oy = o cosθ = 0 cos 53 = 1.04 m / s = o sinθ = 0sin 53 = m / s
15 Example What I know A place kicker kicks a football with a elocity of 0.0 m/s and at an angle of 53 degrees. y = 0 (a) How long is the ball in the air? g = m/s/s =1.04 m/s oy =15.97 m/s What I want to know t =? x =? y max =? y = 1 oyt + gt 0 = (15.97) t 4.9t 15.97t = 4.9t = 4.9t t = 3.6 s
16 Example A place kicker kicks a football with a elocity of 0.0 m/s and at an angle of 53 degrees. (b) How far away does it land? What I know =1.04 m/s oy =15.97 m/s y = 0 g = m/s/s What I want to know t = 3.6 s x =? y max =? x = t (1.04)(3.6) = 39.4 m
17 Example A place kicker kicks a football with a elocity of 0.0 m/s and at an angle of 53 degrees. (c) How high does it trael? CUT YOUR TIME IN HALF! What I know =1.04 m/s oy =15.97 m/s y = 0 g = m/s/s What I want to know t = 3.6 s x = 39.4 m y max =? 1 y = oyt + gt y = (15.97)(1.63) 4.9(1.63) y = m
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